Power Rule (x^n)
Functions of the type $x^n$xn are called power functions, and the power rule is a shortcut way to find the derivative without having to use first principles every time.
But, we will use first principles to determine the rule.
| $f(x)$f(x) | $=$= | $x^n$xn |
| $f(x+h)$f(x+h) | $=$= | $(x+h)^n$(x+h)n |
| $f'(x)$f′(x) | $=$= | $\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)$limh→0(f(x+h)−f(x)h) |
| $f'(x)$f′(x) | $=$= | $\lim_{h\to0}\left(\frac{\left(x+h\right)^n-x^n}{h}\right)$limh→0((x+h)n−xnh) |
From here we need to work through some pretty intense algebra, but it distills down to an understanding of the expansion of $x^n-a^n$xn−an
$x^n-a^n=(x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+a^3x^{n-4}+......+a^{n-1})$xn−an=(x−a)(xn−1+axn−2+a2xn−3+a3xn−4+......+an−1)
Using this expansion we can also determine the expansion for $(x+h)^n-x^n$(x+h)n−xn
$(x+h)^n-x^n==[(x+h)-x][(x+h)^{n-1}+x(x+h)^{n-2}+x^2(x+h)^{n-3}+x^3(x+h)^{n-4}+......+x^{n-1}]$(x+h)n−xn==[(x+h)−x][(x+h)n−1+x(x+h)n−2+x2(x+h)n−3+x3(x+h)n−4+......+xn−1]
| $f'(x)$f′(x) | $=$= |
$\lim_{h\to0}\left(\frac{h\left[\left(x+h\right)^{n-1}+x\left(x+h\right)^{n-2}+x^2\left(x+h\right)^{n-3}+x^3\left(x+h\right)^{n-4}+...x^{n-1}\right]}{h}\right)$limh→0(h[(x+h)n−1+x(x+h)n−2+x2(x+h)n−3+x3(x+h)n−4+...xn−1]h) |
| $f'(x)$f′(x) | $=$= |
$\lim_{h\to0}\left(\left(x+h\right)^{n-1}+x\left(x+h\right)^{n-2}+x^2\left(x+h\right)^{n-3}+x^3\left(x+h\right)^{n-4}+...x^{n-1}\right)$limh→0((x+h)n−1+x(x+h)n−2+x2(x+h)n−3+x3(x+h)n−4+...xn−1) |
| $f'(x)$f′(x) | $=$= |
$x^{n-1}+x\left(x\right)^{n-2}+x^2\left(x\right)^{n-3}+x^3\left(x\right)^{n-4}+...x^{n-1}$xn−1+x(x)n−2+x2(x)n−3+x3(x)n−4+...xn−1 |
| $f'(x)$f′(x) | $=$= |
$x^{n-1}+\frac{x\left(x\right)^n}{x^2}+\frac{x^2\left(x\right)^n}{x^3}+\frac{x^3\left(x\right)^n}{x^4}+...x^{n-1}$xn−1+x(x)nx2+x2(x)nx3+x3(x)nx4+...xn−1 |
| $f'(x)$f′(x) | $=$= |
$x^{n-1}+\frac{x^n}{^x}+\frac{x^n}{^x}+\frac{x^n}{^x}+...x^{n-1}$xn−1+xnx+xnx+xnx+...xn−1 |
| $f'(x)$f′(x) | $=$= |
$x^{n-1}+x^{n-1}+x^{n-1}+x^{n-1}+...x^{n-1}$xn−1+xn−1+xn−1+xn−1+...xn−1 |
| $f'(x)$f′(x) | $=$= | $nx^{n-1}$nxn−1 |
Now of course, that was a lot of algebraic gymnastics, check with your teacher you may or may not need to follow that entire proof. For those that don't here is the key message.
That for functions of the form $f(x)=x^n$f(x)=xn the derivative is $f'(x)=nx^{n-1}$f′(x)=nxn−1 and this is called the power rule.
Examples
Find the derivatives of
a)$f(x)=x^2$f(x)=x2, $f'(x)=2x$f′(x)=2x
b) $g(m)=m^4$g(m)=m4, so $g'(m)=4m^3$g′(m)=4m3
c) $h(t)=t^{\frac{3}{2}}$h(t)=t32, so $h'(t)=\frac{3}{2}t^{\frac{1}{2}}$h′(t)=32t12
Be careful with these fractional ones - they are a common source of student errors.
d) $g(x)=\frac{1}{x^4}$g(x)=1x4, firstly we need the function in power form, so we convert it and get that $g(x)=x^{-4}$g(x)=x−4. Now we can use the power rule and see that $g'(x)=-4x^{-5}$g′(x)=−4x−5 Remember that $-4-1=-5$−4−1=−5. A common mistake here is to arrive at a power of $-3$−3.
For a function $f(x)=x^n$f(x)=xn, the derivative $f'(x)=nx^{n-1}$f′(x)=nxn−1
$n$n can be positive or negative, integer or fraction
Example
Question 1
Consider the function $f\left(x\right)=x^n$f(x)=xn, where $n$n is any positive integer.
Complete the following table. (Use the ^ key to create powers).
$f\left(x\right)$f(x) $f'\left(x\right)$f′(x) $x$x $\editable{}$ $x^2$x2 $\editable{}$ $x^3$x3 $\editable{}$ $x^4$x4 $\editable{}$ $x^n$xn $\editable{}$
Question 2
Consider the function $y=\frac{1}{x^2}$y=1x2.
Rewrite the function in negative index form.
Determine the derivative of $y=\frac{1}{x^2}$y=1x2.
Question 3
Determine the derivative of $y=x^{\frac{6}{5}}$y=x65.